If anyone could answer this question it would be much appreciated. 5^2x-26.5^x+25=0 Thanks
mdm92 New Member Joined Mar 15, 2009 Messages 5 Gender Male HSC 2010 Sep 14, 2009 #1 If anyone could answer this question it would be much appreciated. 5^2x-26.5^x+25=0 Thanks
untouchablecuz Active Member Joined Mar 25, 2008 Messages 1,693 Gender Male HSC 2009 Sep 14, 2009 #2 mdm92 said: If anyone could answer this question it would be much appreciated. 5^2x-26.5^x+25=0 Thanks Click to expand... 5^2x-26.5^x+25=(5^x)^2-26(5^x)+25 let u=5^x u^2-26u+25=0 (u-25)(u-1)=0 u=25 or u=1 but u=5^x .'. 5^x=25=5^5 --> x=5 and 5^x=1=5^0 --> x=0
mdm92 said: If anyone could answer this question it would be much appreciated. 5^2x-26.5^x+25=0 Thanks Click to expand... 5^2x-26.5^x+25=(5^x)^2-26(5^x)+25 let u=5^x u^2-26u+25=0 (u-25)(u-1)=0 u=25 or u=1 but u=5^x .'. 5^x=25=5^5 --> x=5 and 5^x=1=5^0 --> x=0
K khorne Guest Sep 14, 2009 #3 untouchablecuz said: 5^2x-26.5^x+25=(5^x)^2-26(5^x)+25 let u=5^x u^2-26u+25=0 (u-25)(u-1)=0 u=25 or u=1 but u=5^x .'. 5^x=25=5^5 --> x=5 and 5^x=1=5^0 --> x=0 Click to expand... 5^x = 25...x = 2
untouchablecuz said: 5^2x-26.5^x+25=(5^x)^2-26(5^x)+25 let u=5^x u^2-26u+25=0 (u-25)(u-1)=0 u=25 or u=1 but u=5^x .'. 5^x=25=5^5 --> x=5 and 5^x=1=5^0 --> x=0 Click to expand... 5^x = 25...x = 2